In backlighting applications, discharge lamps (or fluorescent lamps) have recently been increasingly substituted with LED lighting modules. Such lighting modules may comprise several printed circuit boards (PCBs), each of them with one or more LEDs, the PCBs being connected through wires or cables in order to be flexible and fit into channel letter light signs, that normally require custom light sources, or adaptable systems. The high number of units (or modules) connected in parallel requires a low cost driving solution.
The linear current regulation device according to the diagram of FIG. 1 is an example of a solution presently used for the purpose.
In this example, that refers to the driving of two LEDs connected in series and respectively indicated by LED1 and LED2, the device includes two bipolar transistors (BJT), in this case of a pnp type, indicated by the reference numbers T1 and T2, and two resistances R1 and R2.
Referring to FIG. 1, the positive terminal of a direct voltage input generator VDC1 is connected to the emitter E1 of the first transistor T1 (node A). The resistance R1 is connected between the emitter E1 and the base B1 of the first transistor T1. The emitter E2 of the second transistor T2 is connected (node C) to the base B1. The collector C1 of the first transistor T1 is connected, via the node D, to the base B2 of the second transistor T2. The resistance R2 is connected between the node D and the node B, i.e. the negative terminal of the voltage generator VDC1. The collector C2 of the second transistor T2 feeds the series of both LED1 and LED2. Finally, the cathode of the second diode LED2 is connected to the above mentioned node B on the second supply line 20.
In operation, the high-impedance resistance R2 polarizes the first transistor T1 with a very low collector current Ic (μA), but the base-emitter voltage VBE1 of the first transistor T1 is set to the value VBEon.
The current ILED that flows through and drives the LEDs is the same as the collector current IC2 of the transistor T2 (which approximately equals the emitter current IE2 of the same transistor) and therefore it is dependent, through the resistance R1, on the value of the voltage dropping between the emitter and the base (reference voltage VBE1) of the transistor T1, according to the following relation:
      I    LED    =                    I                  C          ⁢                                          ⁢          2                    ≈              I                  E          ⁢                                          ⁢          2                      =                  V                  BE          ⁢                                          ⁢          1                            R        1            
In the driver device of FIG. 1, the transistor T2 is therefore used as a driver circuit of the LEDs, while the transistor T1 has a stabilizing function.
The collector current IC2 of the driver transistor T2 drives the light sources LED1 and LED 2 and the stabilization circuit T1, R1, R2 generates a reference voltage VBE1—stable with reference to the input voltage VDC1—which, applied to the transistor T2, makes the current ILED flowing through the LEDs equally stable with reference to the input voltage VDC1. The consequence is a stabilization of the current ILED with reference to the input voltage VDC1.
Nevertheless, there remains a dependence of the current ILED from the usage temperature T (because of the thermal drifts of VBE).
A possible range of the operating temperatures of LED modules goes from −30° C. to +80° C., because of the different ambient conditions where they can be employed.
The base-emitter junction voltage of the transistor T1 varies with the temperature T on the basis of a coefficient k (mV/° C.).
Considering the reference standard ambient temperature of 25° C., the current ILED that flows in the LEDs varies on the basis of the following relation:
                                          I            LED                    ⁡                      (            T            )                          =                ⁢                                                            V                                  BE                  ⁢                                                                          ⁢                  1                                            ⁡                              (                T                )                                                    R              1                                =                                                                                          V                                          BE                      ⁢                                                                                          ⁢                      1                                                        ⁡                                      (                    T0                    )                                                  +                                                      k                    ·                    Δ                                    ⁢                                                                          ⁢                  T                                                            R                1                                      =                                                                                V                                          BE                      ⁢                                                                                          ⁢                      1                                                        ⁡                                      (                    T0                    )                                                  +                                  k                  ⁡                                      (                                          T                      -                                              T                        ⁢                                                                                                  ⁢                        0                                                              )                                                                              R                1                                                                            =                ⁢                                                                              V                                      BE                    ⁢                                                                                  ⁢                    1                                                  ⁡                                  (                                      T                    ⁢                                                                                  ⁢                    0                                    )                                                            R                1                                      +                                          k                ⁡                                  (                                      T                    -                                          T                      ⁢                                                                                          ⁢                      0                                                        )                                                            R                1                                              =                                                    I                LED                            ⁡                              (                                  T                  ⁢                                                                          ⁢                  0                                )                                      +                          Δ              ⁢                                                          ⁢                                                I                  LED                                ⁡                                  (                  T                  )                                                                        
The value of the current ILED(T) at a given temperature T is therefore given by the value of the current ILED(T0) flowing in the LEDs at the reference temperature of 25° C. with a variation ΔILED(T) (positive or negative) which depends on the temperature T.
We can define:
      H    ⁢                  ⁢    1    =                    ⅆ                  ⅆ          T                    ⁢              (                                            I              LED                        ⁡                          (              T              )                                                          I              LED                        ⁡                          (                              T                ⁢                                                                  ⁢                0                            )                                      )              =          k              V                  BE          ⁢                                          ⁢          1                    
A typical value for the coefficient k for a p-n junction is −2 mV/° C.
Let us assume a numerical example where:
ILED(25° C.)=30 mA,
VBE1(25° C.)=0.6V,
R1=20Ω.
In this case, the current at a low temperature of −30° C. increases up to the value of:
                                          I            LED                    ⁡                      (                                          -                30                            ⁢              °              ⁢                                                          ⁢              C                        )                          =                                                            V                                  BE                  ⁢                                                                          ⁢                  1                                            ⁡                              (                                  25                  ⁢                  °                  ⁢                                                                          ⁢                  C                                )                                                    R              1                                +                                    k              ⁡                              (                                                      -                    30                                    -                  25                                )                                                    R              1                                                              =                              30            ⁢                                                  ⁢            mA                    +                                                                      -                  0.002                                ⁢                                  (                                                            -                      30                                        -                    25                                    )                                            20                        ⁢            mA                                                  =                  35.5          ⁢                                          ⁢          mA                    while at a high temperature of +80° C. the current decreases to a value of:
                                          I            LED                    ⁡                      (                          80              ⁢              °              ⁢                                                          ⁢              C                        )                          =                                                            V                                  BE                  ⁢                                                                          ⁢                  1                                            ⁡                              (                                  25                  ⁢                  °                  ⁢                                                                          ⁢                  C                                )                                                    R              1                                +                                    k              ⁡                              (                                  80                  -                  25                                )                                                    R              1                                                              =                              30            ⁢                                                  ⁢            mA                    +                                                                      -                  0.002                                ⁢                                  (                                      80                    -                    25                                    )                                            20                        ⁢            mA                                                  =                  24.5          ⁢                                          ⁢          mA                    
In conclusion, it can be stated that the current at a temperature of −30° C. amounts approximately to 118% of the current at ambient temperature (ILED(−30° C.)=118% ILED(25° C.)), while at a temperature of +80° C. it amounts approximately to 82% of the current at ambient temperature (ILED(80° C.)=82% ILED(25° C.)).
Consequently, the LED module input power could have a variation of ±20% from the rated input power at a reference ambient temperature of 25° C.